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In deriving the existence criterion of $\operatorname{Spin}_{\mathbb{C}}$ structure Ralph Cohen in his notes on the topology of fiber bundle (pp.169) uses the following diagram $\require{AMScd}$ \begin{CD} \operatorname{BSpin}_{\mathbb{C}}(n) @>c>> K(\mathbb{Z},2) \\ @VfVV @VVpV \\ \operatorname{BSO}(n) @>w_2>> K(\mathbb{Z}_2,2) \end{CD} where

$f: \operatorname{BSpin}_{\mathbb{C}}(n) \to \operatorname{BSO}(n)$ follows from the $U(1)$-bundle $\operatorname{Spin}_{\mathbb{C}}(n) \to \operatorname{Spin}(n)/\mathbb{Z}_2 = \operatorname{SO}(n)$

$p: K(\mathbb{Z},2) \to K(\mathbb{Z}_2,2)$ follows from the projection $\mathbb{Z} \to \mathbb{Z}_2$

$c: \operatorname{BSpin}_{\mathbb{C}}(n) \to K(\mathbb{Z},2)$ follows from the projection map $\operatorname{Spin}_{\mathbb{C}}(n) \to U(1)/\mathbb{Z}_2 = U(1)$

$w_2: \operatorname{BSO}(n) \to K(\mathbb{Z}_2,2)$ follows from applying the cohomology classification via Eilenberg-Maclane spaces to $w_2 \in H^2(\operatorname{BSO}(n);\mathbb{Z}_2)$

Cohen claims that by converting $p$ and $w_2$ to Serre fibrations the diagram implies $\operatorname{BSpin}_{\mathbb{C}}(n)$ is homotopy equivalent to the pullback along $w_2$ of $K(\mathbb{Z},2)$ which I can't see. Could somebody explain why it is true?

Alex Provost
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PhysicsMath
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1 Answers1

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Consider the general setting in which we have a square \begin{array}{ccc} W & \rightarrow & X \\ \downarrow & & \downarrow \\ Y & \rightarrow & Z \end{array}

In any case, we have a map $W \to X \times_Z Y$. To show that this is an equivalence, it suffices to show that its (homotopy) fiber is trivial. (There's a name for this kind of limit: it's called the total homotopy fiber of the square, and it's a measure of the "cartesianness" of the square.)

The fiber of $W \to X \times_Z Y$ can be computed iteratively as the fiber of the induced map of fibers $$\operatorname{fib}(W \to X) \to \operatorname{fib}(Y \to Z).$$

Now what do you know about the fibers of the rows in your particular square?

JHF
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    Many thanks for the answer! This seems to be a very general theorem that should be covered in standard textbook (unfortunately I did not find it in Hatcher). I am not very clear about the two key statements in your answer: (1) do you mean that in general to show $X$ and $Y$ are homotopy equivalent it suffices to show $X \twoheadrightarrow Y$ has trivial (homotopy) fiber? I actually can't see why (by reducing it to the definition of homotopy equivalence) (2) the fiber of $W \to X \times_z Y$ can be computed iteratively as the fiber of the induced map of fibers what do you mean by "iteratively"? – PhysicsMath Aug 08 '16 at 14:17
  • For (1) do you mean the fiber wise weak homotopy equivalence $\Leftrightarrow$ homotpy Cartesian i.e. $W$ is weak homotopy equivalent to $X \times_Z Y$? – PhysicsMath Aug 08 '16 at 17:21
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    Suppose $X \to Y$ has trivial homotopy fiber. Then the long exact sequence in homotopy shows the map $X \to Y$ is a weak homotopy equivalence. For nice spaces, this is the same as a homotopy equivalence. In this special case, I'm proposing that you can show that the map $W \to X \times_Z Y$ has trivial homotopy fiber, so the square is homotopy cartesian. Secondly, by iteratively I mean $\operatorname{fib}(W \to X \times_Z Y) \simeq \operatorname{fib}(\operatorname{fib}(W \to X) \to \operatorname{fib}(Y \to Z))$. – JHF Aug 09 '16 at 01:05
  • I see. So essentially the homotopy equivalence follows from the fact that the commutative square is a homotopy Cartesian. And to show that it is a homotopy Cartesian we show $W \to X \times_Z Y$ has trivial homotopy fiber. – PhysicsMath Aug 09 '16 at 03:02
  • I actually am not aware of $\operatorname{fib}(W \to X \times_Z Y) \simeq \operatorname{fib}(\operatorname{fib}(W \to X) \to \operatorname{fib}(Y \to Z))$. Could you add this up to the answer and also the goal of proving homotopy Cartesian? This would help people understand your answer at a glance (at-a-glance understanding is advocated by George Poyle in his "how to solve it"). – PhysicsMath Aug 09 '16 at 03:05